深度优先遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
vector<vector<int>> res;
public:
void dfs(TreeNode* root, vector<int>& cur, int targetSum)
{
if(!root)
return;
targetSum -= root->val;
cur.push_back(root->val);
// 如果是叶子节点并且符合目标值,则存储
// 注意,这里不能直接返回,因为push的元素还没有pop,会导致回溯出问题
if(root->left == nullptr && root->right == nullptr && targetSum == 0)
res.push_back(cur);
dfs(root->left, cur, targetSum);
dfs(root->right, cur, targetSum);
cur.pop_back();
}
vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
vector<int> cur;
dfs(root, cur, targetSum);
return res;
}
};